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<h3 class="heading"><span class="type">Paragraph</span></h3>
<p>For the second linear independent solution, there are three cases:Case 1 <span class="process-math">\(r_1-r_2 \neq \mathbb{Z}\text{,}\)</span></p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq5_18.html ./knowl/eq5_19.html ./knowl/eq5_11.html">
\begin{equation*}
y_2=x^{r_2} \sum_{n=0}^{\infty} b_n x^n \quad \textrm{with}~b_0=1 \quad \textrm{for}~0&lt;x&lt;\rho.
\end{equation*}
</div>
<p class="continuation">Case 2 <span class="process-math">\(r_1-r_2=0\text{,}\)</span></p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq5_18.html ./knowl/eq5_19.html ./knowl/eq5_11.html">
\begin{equation*}
y_2=y_1(x) \ln x+x^{r_1} \sum_{n=1}^{\infty} c_n x^n,\quad \textrm{for}~0&lt;x&lt;\rho.
\end{equation*}
</div>
<p class="continuation">Case 3 <span class="process-math">\(r_1-r_2=\mathbb{Z}\text{,}\)</span></p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq5_18.html ./knowl/eq5_19.html ./knowl/eq5_11.html">
\begin{equation}
y_2(x)=a y_1(x) \ln x+x^{r_2} \sum_{n=0}^{\infty} d_n x^n \quad \textrm{with}~d_0=1 \quad \textrm{for}~0&lt;x&lt;\rho.\tag{5.5.9}
\end{equation}
</div>
<p class="continuation">In (<a href="" class="xref" data-knowl="./knowl/eq5_18.html" title="Equation 5.5.8">(5.5.8)</a>) to (<a href="" class="xref" data-knowl="./knowl/eq5_19.html" title="Equation 5.5.9">(5.5.9)</a>), <span class="process-math">\(a_n, b_n, c_n, d_n,  (n \geq 1)\)</span> and <span class="process-math">\(a\)</span> can be determined by substituting the solution into (<a href="" class="xref" data-knowl="./knowl/eq5_11.html" title="Equation 5.5.1">(5.5.1)</a>).</p>
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